Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\sqrt{x^2-\frac{1}{4}-\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\) (ĐK: \(x\ge\frac{-1}{2}\) )
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}-\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left[2x\left(x^2+1\right)+\left(x^2+1\right)\right]\)
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}-x-\frac{1}{2}}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow\sqrt{\left(x+\frac{1}{2}\right)^2}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow2x+1=\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow\left(x^2+1\right)\left(2x+1\right)-\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x^2+1-1\right)=0\)
\(\Leftrightarrow x^2\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=0\end{cases}}\) (nhận)
Vậy .....
\(\sqrt{x^2-\frac{1}{4}-\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}-\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left[x^2\left(2x+1\right)+2x+1\right]\)
\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}-\left|x+\frac{1}{2}\right|}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)(1)
Vì VT > 0 nên VP >0
\(\Leftrightarrow\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\ge0\)
\(\Leftrightarrow x\ge-\frac{1}{2}\)
Khi đó \(\left(1\right)\Leftrightarrow\sqrt{x^2-\frac{1}{4}-x-\frac{1}{2}}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow\sqrt{x^2-x-\frac{3}{4}}=\frac{1}{2}\left(x^2+1\right)\left(2x+1\right)\)
\(\Leftrightarrow x^2-x-\frac{3}{4}=\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)^2\)
\(\Leftrightarrow\left(2x-3\right)\left(2x+1\right)-\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-3-\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\2x-3=\frac{1}{4}\left(x^2+1\right)^2\left(2x+1\right)\end{cases}}\)
Cần cù bù thông minh , phá tung pt dưới ra được cái phương trình bậc 5, sau đó dùng Wolfram|Alpha: Computational Intelligence để tính nghiệm rồi phân tích nhân tử =))
\(DKXD:x>0\)
\(PT\Leftrightarrow\sqrt{x+\frac{3}{x}}-2=\frac{x^2+7}{2\left(x+1\right)}-2\)
\(\Leftrightarrow\frac{x+\frac{3}{x}-4}{\sqrt{x+\frac{3}{x}}+2}=\frac{x^2-4x-4+7}{2\left(x+1\right)}\)
\(\Leftrightarrow\frac{x^2-4x+3}{x\sqrt{x+\frac{3}{x}}+2x}-\frac{x^2-4x+3}{2\left(x+1\right)}=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(\frac{1}{x\sqrt{x+\frac{3}{x}}+2x}-\frac{1}{2\left(x+1\right)}\right)=0\)
\(\Rightarrow x=1\text{ }or\text{ }x=3\text{ }or\text{ }x\sqrt{x+\frac{3}{x}}=2\text{ }\)
\(\Leftrightarrow x=1\text{ }or\text{ }x=3\text{ }or\text{ }x^3+3x-4=0\)
\(\Leftrightarrow x=1\text{ }or\text{ }x=3\text{ }or\text{ }x^3+3x-4=0\)
\(\Leftrightarrow x=1\text{ }or\text{ }x=3\text{ }or\left(\text{ }x-1\right)\left(x^2+x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)
Vậy PT có 2 nghiệm \(x=1;x=3\)
\(\frac{15}{2}\left(30x^2-4x\right)=2004\left(\sqrt{30060x+1}+1\right)\)
\(\Leftrightarrow5\left(15x^2-2x\right)=668\left(\sqrt{30060x+1}+1\right)\)
\(\Leftrightarrow75x^2-10x-1340008=668\left(\sqrt{30060x+1}-2005\right)\)
\(\Leftrightarrow\left(5x+668\right)\left(15x-2006\right)=\frac{1338672\left(15x-2006\right)}{\left(\sqrt{30060x+1}+2005\right)}\)
\(\Leftrightarrow\left(15x-2006\right)\left(5x+668-\frac{1338672}{\left(\sqrt{30060x+1}+2005\right)}\right)=0\)
Tới đây tự làm tiếp nhá.
câu này chỉ cần đưa ề đối xúng là được thôi
\(\Leftrightarrow\left(15x\right)^2-30x=2004\sqrt{30060x+1}+2004\)
\(\Leftrightarrow\left(15x-1\right)^2=2004\sqrt{30060x+1}+2005\)
đặt \(\sqrt{30060x+1}=15y-1\)
\(\Leftrightarrow\hept{\begin{cases}\left(15x-1\right)^2=2004\left(15y-1\right)+2005\\\left(15y-1\right)^2=30060x+1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(15x-1\right)^2=30060y+1\\\left(15y-1\right)^2=30060x+1\end{cases}}\)
đến đây thì lấy cái đầu trừ cái thứ 2 là ra
\(DK:x\ge0\)
\(\Leftrightarrow\frac{\sqrt{x}-\sqrt{x+1}}{x-x-1}+\frac{\sqrt{x+1}-\sqrt{x+2}}{x+1-x-2}+\frac{\sqrt{x+2}-\sqrt{x+3}}{x+2-x-3}=1\)
\(\Leftrightarrow-\sqrt{x}+\sqrt{x+1}-\sqrt{x+1}+\sqrt{x+2}-\sqrt{x+2}+\sqrt{x+3}=1\)
\(\Leftrightarrow\sqrt{x+3}-\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x+3}=1+\sqrt{x}\)
\(\Leftrightarrow x+3=x+2\sqrt{x}+1\)
\(\Leftrightarrow x=1\)
Vay nghiem cua PT la \(x=1\)
\(18x^2-2x-\frac{17}{3}+9\sqrt{x-\frac{1}{3}}=0\)
Điều kiện: \(x\ge\frac{1}{3}\)
Đặt \(\sqrt{x-\frac{1}{3}}=a\left(a\ge0\right)\)
\(\Rightarrow x=a^2+\frac{1}{3}\)
Ta suy ra phương trình tương đương với
\(18\left(a^2+\frac{1}{3}\right)^2-2\left(a^2+\frac{1}{3}\right)-\frac{17}{3}+9a=0\)
\(\Leftrightarrow54a^4+30a^2+27a-13=0\)
\(\Leftrightarrow\left(3a-1\right)\left(18a^3+6a^2+12a+13\right)=0\)
Dễ thấy \(18a^3+6a^2+12a+13>0\) vì \(a\ge0\)
\(\Rightarrow3a-1=0\)
\(\Leftrightarrow a=\frac{1}{3}\)
\(\Leftrightarrow\sqrt{x-\frac{1}{3}}=\frac{1}{3}\)
\(\Leftrightarrow x-\frac{1}{3}=\frac{1}{9}\)
\(\Leftrightarrow x=\frac{4}{9}\)
\(x=\frac{1}{2}\left(\sqrt{2}-1\right)\)
\(\Leftrightarrow2x=\sqrt{2}-1\Leftrightarrow4x^2=3-2\sqrt{2}=1-4.\frac{1}{2}\left(\sqrt{2}-1\right)=1-4x\)
\(\Leftrightarrow4x^2+4x-1=0\)
\(\left[x^3\left(4x^2+4x-1\right)+1\right]^{19}=1^{19}=1\)
\(\sqrt{x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+4x^2+4x-1+4}^3=\sqrt{4}^3=8\)
\(\frac{1-\sqrt{2}x}{\sqrt{\frac{1}{2}\left(4x^2+4x-1\right)+\frac{1}{2}}}=\frac{1-\sqrt{2}x}{\sqrt{\frac{1}{2}}}=\sqrt{2}-2x=\sqrt{2}-\left(\sqrt{2}-1\right)=1\)
\(M=1+8+1=10\)
gõ lại đề